I'm trying to tackle the following question:

Prove that the equation $x^2+y^2-z^2+xz-yz-1=0$ determines $z$ as a function of $x,y$ near $(1,0,1)$ and find $\displaystyle \frac{\partial z}{\partial x}(1,0),\frac{\partial z}{\partial y}(1,0),\frac{\partial^2 z}{\partial x \partial y}$.

My try:

Let $f(x,y,z)=x^2+y^2-z^2+xz-yz-1$, such that $(1,0,1)$ is a root of $f$.

Assume that exists $z(x,y)\in C^1$ near $(x,y)=(1,0)$ such that $f(x,y,z(x,y))=0$.

Derive wrt $x$: $\displaystyle \frac{\partial f}{\partial x}+\frac{\partial f}{\partial z}\cdot\frac{\partial z}{\partial x}=0$.

Derive wrt $y$: $\displaystyle \frac{\partial f}{\partial y}+\frac{\partial f}{\partial z}\cdot\frac{\partial z}{\partial y}=0$.

So, $\displaystyle \frac{\partial f}{\partial z}\begin{pmatrix}\frac{\partial z}{\partial x} \\ \frac{\partial z}{\partial y}\end{pmatrix}=-\begin{pmatrix}\frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y}\end{pmatrix}$. Now,$$\left.\frac{\partial f}{\partial z}\right\vert_{(1,0,1)}=\Big.(x-2z-y)\Big\vert_{(1,0,1)}=-1 \\ \left.\frac{\partial f}{\partial x}\right\vert_{(1,0,1)}=\Big.(2x+z)\Big\vert_{(1,0,1)}=3 \\ \left.\frac{\partial f}{\partial y}\right\vert_{(1,0,1)}=\Big.(2y-z)\Big\vert_{(1,0,1)}=-1$$Hence, $\displaystyle -1\left.\begin{pmatrix}\frac{\partial z}{\partial x} \\ \frac{\partial z}{\partial y} \end{pmatrix}\right\vert_{(1,0,1)}=-\begin{pmatrix}3 \\ -1\end{pmatrix} \Rightarrow \frac{\partial z}{\partial x}(1,0)=3,\frac{\partial z}{\partial y}(1,0)=-1$.

Unfortunately,I don't know how to compute $\displaystyle \frac{\partial^2 z}{\partial x \partial y}(1,0)$.

Please help, thank you!

  • $\begingroup$ $x^2+y^2-z^2+xy-yz-1=0$ is not satisfied at $(1,0,1)$ - what's going on with the "-1" part ? $\endgroup$
    – WW1
    Dec 7 '15 at 21:30
  • $\begingroup$ @WW1, you are right, it should be $xz$ instead of $xy$. I will edit my post. $\endgroup$
    – Galc127
    Dec 7 '15 at 21:32

Derive once with respect to $x$ and then once with respect to $y$:

$$0=\partial_x(\partial_y(x^2+y^2-z^2+xz-yz-1))=\partial_x(2y-2z\partial_y z+x\partial_yz-y\partial_yz-z )=-2\partial_xz\,\partial_y z-2z\,\partial_x\partial_y z+x\partial_x\partial_yz+\partial_yz-y\,\partial_x\partial_yz-\partial_xz .$$

Now you know $x$, $y$, $z$, $\partial_xz $, and $\partial_yz$, hence you can find $\partial_x\partial_yz$.

  • $\begingroup$ Thank you, I got $\displaystyle \frac{\partial^2 z}{\partial x \partial y}=-1$. Is that correct? $\endgroup$
    – Galc127
    Dec 7 '15 at 21:53

Thank you for the edit, I was really struggling with this earlier. Assume that z is a function of x and y.

Derivative with r.w.t x:

$$2x - 2z\frac{\partial z}{\partial x} + z + x\frac{\partial z}{\partial x} - y\frac{\partial z}{\partial x} = 0$$

Solving that for $\frac{\partial z}{\partial x}$:

$$\frac{\partial z}{\partial x} = -\frac{2x + z}{x - 2z - y}$$

$$\frac{\partial z}{\partial x}(1, 0) = 3$$

Follow that same process for each derivative.


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